Equilibrium pressures question?
If #5.00*10^-1# atm of #HI# gas, #0.00100# atm of #H_2# gas, and #0.00500# atm of #I_2# gas are present in a #5.00# L flask, calculate the equilibrium pressures of all 3 gases once equilibrium is established ?
If
2 Answers
#P_(H_2) = "0.0402 atm"#
#P_(I_2,eq) = "0.0442 atm"#
#P_(HI,eq) = "0.4216 atm"#
Well, from your notes, you seem to have
#"H"_2(g) " "" "+" "" " "I"_2(g)" " rightleftharpoons " " 2"HI"(g)#
#"I"" "0.00100" "" "" "0.00500" "" "" "0.500#
#"C"" "+x" "" "" "" "+x" "" "" "" "-2x#
#"E"" "0.00100+x" "0.00500+x" "0.500-2x#
Here, we have written that the reaction goes in reverse. Now that you have
#K_p = 100 = P_(HI)^2/(P_(H_2)P_(I_2))#
#= (0.500 - 2x)^2/((0.00100 + x)(0.00500 + x))#
Not much we can do here other than solve it in full;
#100 = (0.500 - 2x)^2/(5 xx 10^(-6) + 0.00600x + x^2)#
Multiply through by the denominator.
#5 xx 10^(-4) + 0.600x + 100x^2 = (0.500 - 2x)^2#
Expand the right-hand side.
#5 xx 10^(-4) + 0.600x + 100x^2 = 0.250 - 2 cdot 0.500 cdot 2x + 4x^2#
Get this into standard quadratic form:
#-0.2495 + 0.600x + 100x^2 = -2x + 4x^2#
#96x^2 +2.600x - 0.2495 = 0#
Now,
#a = 96#
#b= 2.600#
#c = -0.2495#
so that
#x = (-b pm sqrt(b^2 - 4ac))/(2a)#
#= (-(2.600) pm sqrt(2.600^2 - 4(96)(-0.2495)))/(2(96))#
This, as-given, has:
#x = 0.0392, -0.0662#
When we plug these in, however, only
#color(blue)(P_(H_2) = 0.00100 + (0.0392) = "0.0402 atm")#
#color(blue)(P_(I_2,eq) = 0.00500 + (0.0392) = "0.0442 atm")#
#color(blue)(P_(HI,eq) = 0.500 - 2(0.0392) = "0.4216 atm")#
use an ICE table, then find pressures with
Explanation:
My notes
Classmate's notes